Problem integrating $\int\frac{1}{\sqrt[3]{x}+\sqrt[4]{x}}dx$

Hint: take $u=\sqrt[12]{x} $. The integral becomes (why?) $$\int\frac{1}{\sqrt[3]{x}+\sqrt[4]{x}}dx=12\int\frac{u^{8}}{u+1}du $$ and now doing long division we get $$12\int\frac{u^{8}}{u+1}du=12\int\left(u^{7}-u^{6}+u^{5}-u^{4}+u^{3}-u^{2}+u-1+\frac{1}{u+1}\right)du.$$

$$=\int \frac{1}{x^{1/3}+x^{1/4}}dx$$

Factor out a $x^{1/4}$:

$$=\int \frac{1}{x^{1/4}(x^{1/12}+1)}dx$$

Let $u=x^{1/12}$. Then $du=\frac{1}{12} x^{-11/12} \,dx$ and $12x^{11/12}\,du=12u^{11}\,du=dx$. Also $x^{1/4}=u^3$. So we have:

$$=12 \int \frac{u^{11}}{u^3(u+1)}\,du$$

$$=12 \int \frac{u^8}{u+1}\, du$$

This is standard with another substitution $y=u+1$. The binomial theorem might be of great help.